The efficiency of a Carnot engine is 25%. The efficiency is increased to 30% when the sink temperature is reduced by 20°C. What will be the source temperature?

Concept:

The efficiency of Carnot or any reversible engine is given by:

 \({\eta _{Carnot}} = {\eta _{Rev}} = 1 - \frac{{{T_2}}}{{{T_1}}}\)

where, η = Efficiency of Carnot Engine, T₁ = Source temperature in K and T₂ = Sink Temperature in K.

Calculation:

Given:

η_I = 0.25 when T₁ = Source temperature (K) and T₂ = Sink Temperature (K)

η_II = 0.30 when  T₁ = Source temperature(K) and (T₂)_new = Sink Temperature (K); (T₂)_new = T₂ – 20

\({\eta _I} = 1 - \frac{{{T_2}}}{{{T_1}}}\)

\( \Rightarrow 0.25 = 1 - \frac{{{T_2}}}{{{T_1}}}\)

\( \Rightarrow \frac{{{T_2}}}{{{T_1}}} = 0.75\;\)  

⇒ T₂ = 0.75T₁    …….eqn (i)

Now,

\({\eta _{II}} = 1 - \frac{{{{\left( {{T_2}} \right)}_{new}}}}{{{T_1}}}\)

\( 0.3 = 1 - \frac{{{T_2} - 20}}{{{T_1}}}\)

\(\frac{{{T_2} - 20}}{{{T_1}}} = 0.7\)    ……eq(ii)

Putting values from eq(i) in eq(ii),

\(\frac{{0.75{T_1} - 20}}{{{T_1}}} = 0.7\)

⇒ 0.05T₁ = 20

⇒ T₁ = 400 K

∴ The source temperature is 400 K.

Mistake Points

1) Temperature in the efficiency formula is only taken in terms of Kelvin.

2) The difference in degree celsius scale is the same as the Kelvin scale.